The problem states as follows.
Let $S$ be a set of non-zero polynomials over a field $ F$. If there
are no two polynomials of the same degree in $S$, then $S$ is linearly
independent.
I tried the following.
Let be $f_1,\ldots,f_n$ be polynomials chosen such that if $i<j$ then $\deg f_i < \deg f_j$.
If we put $$c_1f_1+\ldots+c_nf_n=0$$
then $$c_nf_n=c_1f_1+\ldots+c_{n-1}f_{n-1}.$$If $c_n\neq 0$, then $c_nf_n\neq 0$ so $c_1f_1+\ldots+c_{n-1}f_{n-1}\neq 0$ but we know that $$\begin{align}\deg(c_nf_n)&=\deg (c_1f_1+\ldots+c_{n-1}f_{n-1})\\&\leq \max \{\deg f_1, \ldots,\deg f_{n-1}\}\\&=\deg f_{n-1}\\&<\deg f_n\end{align}$$ and this is a contradiction. Then we have that $c_n=0$.
By applying the same process, we found that $c_1=\ldots=c_{n-1}=0$.
So $S$ is linearly independent.
Is my proof correct? Is there any shortest one?
Thanks for your help.
Best Answer
Your proof is fine, I suppose. A minor reformulation if one wants to avoid the slightly informal "repeat the process" and "$\ldots$" could be to either write this as proof by induction ($c_n\ne 0$ leads to a contradiction; hence $c_n=0$. But then by induction hypotheses also $c_k=0$ for $1\le k\le n-1$). Or assume that not all $c_i$ are $=0$, let $m$ be maximal with $c_m=0$, then write $f_M$ in terms of the lower degree $f_i$ and arrive at the same contradiction.